first the data, then my question:
df <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
df
Z Y A
1 2 1
2 2 2
3 2 3
1 3 4
2 3 5
3 3 6
I am using R and would like to generate a vector a with given vector Z and Y. Like if (z1==1 & y1==2), then a1=1, if (z3==3 & y1==2), then a1=3....If we only have one condition (Z or Y), I could generate A with switch function, however, how could I generate A with Z and Y?
If I understood the question correctly the answer is as follows. If I misunderstood then please correct me.
df <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
df$A <- NA # Empty out A
# Re-create A
df$A[df$Z == 1 & df$Y == 2] <- 1
df$A[df$Z == 2 & df$Y == 2] <- 2
df$A[df$Z == 3 & df$Y == 2] <- 3
df$A[df$Z == 1 & df$Y == 3] <- 4
df$A[df$Z == 2 & df$Y == 3] <- 5
df$A[df$Z == 3 & df$Y == 3] <- 6
df <- df[order(df$A),]
df
Z Y A
1 1 2 1
5 2 2 2
3 3 2 3
4 1 3 4
2 2 3 5
6 3 3 6
Another way to do it in one call is to use ifelse function as follows:
df <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
df$A <- ifelse(df$Z == 1 & df$Y == 2,1,
ifelse(df$Z == 3 & df$Y == 2,3,
ifelse(df$Z > 1,5,8)))
Z Y A
1 1 2 1
2 2 3 5
3 3 2 3
4 1 3 8
5 2 2 5
6 3 3 5
As the number of conditions increases, if you're just looking at exact matches, it becomes more useful to use a lookup table for the results and merge them. So we start with a lookup table as you've defined it
df.lookup <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
And let's say that we have a dataframe we want to populate
set.seed(17) # To get repeatable random numbers
df <- data.frame(Z=sample(1:3, 6, replace=T), Y=sample(2:3, 6, replace=T))
So we have
> df
Z Y
1 1 2
2 3 2
3 2 3
4 3 2
5 2 2
6 2 2
> merge(df, df.lookup)
Z Y A
1 1 2 1
2 2 2 5
3 2 2 5
4 2 3 2
5 3 2 3
6 3 2 3
The merge function will automatically select same-named columns, but you can specify the names if they are different, see ?merge for more information.
Related
Suppose I have a dataframe that looks like this:
> data <- data.frame(x = c(1,1,2,2,3,4,5,6), y = c(1,2,3,4,5,6,7,8))
> data
x y
1 1 1
2 1 2
3 2 3
4 2 4
5 3 5
6 4 6
7 5 7
8 6 8
I want to use mutate and case_when to create a new id variable that will identify rows using the variable x, and give rows missing x a unique id. In other words, I should have the same id for rows one and two, rows three and four, while rows 5-8 should have their own unique ids. Suppose I want to generate these id values with a function:
id_function <- function(x, n){
set.seed(x)
res <- character(n)
for(i in seq(n)){
res[i] <- paste0(sample(c(letters, LETTERS, 0:9), 32), collapse="")
}
res
}
id_function(1, 1)
[1] "4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf"
I am trying to use this function on the RHS of a case_when expression like this:
data %>%
mutate(my_id = id_function(1234, nrow(.)),
my_id = dplyr::case_when(!is.na(x) ~ id_function(x, 1),
TRUE ~ my_id))
But the RHS does not seem to be vectorized and I get the same value for all non-missing values of x:
x y my_id
1 1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
3 2 3 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
4 2 4 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
5 NA 5 0vnws5giVNIzp86BHKuOZ9ch4dtL3Fqy
6 NA 6 IbKU6DjvW9ypitl7qc25Lr4sOwEfghdk
7 NA 7 8oqQMPx6IrkGhXv4KlUtYfcJ5Z1RCaDy
8 NA 8 BRsjumlCEGS6v4ANrw1bxLynOKkF90ao
I'm sure there's a way to vectorize the RHS, what am I doing wrong? Is there an easier approach to solving this problem?
I guess rowwise() would do the trick:
data %>%
rowwise() %>%
mutate(my_id = id_function(x, 1))
x y my_id
1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 3 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
2 4 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
3 5 e5lMJNQEhtj4VY1KbCR9WUiPrpy7vfXo
4 6 3kYcgR7109DLbxatQIAKXFeovN8pnuUV
5 7 bQ4ok7OuDgscLUlpzKAivBj2T3m6wrWy
6 8 0jSn3Jcb2HDA5uhvG8g1ytsmRpl6CQWN
purrr map functions can be used for non-vectorized functions. The following will give you a similar result. map2 will take the two arguments expected by your id_function.
library(tidyverse)
data %>%
mutate(my_id = map2(x, 1, id_function))
Output
x y my_id
1 1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
3 2 3 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
4 2 4 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
5 3 5 e5lMJNQEhtj4VY1KbCR9WUiPrpy7vfXo
6 4 6 3kYcgR7109DLbxatQIAKXFeovN8pnuUV
7 5 7 bQ4ok7OuDgscLUlpzKAivBj2T3m6wrWy
8 6 8 0jSn3Jcb2HDA5uhvG8g1ytsmRpl6CQWN
I'm not very experienced R user, so seek advice how to optimize what I've build and in which direction to move on.
I have one reference data frame, it contains four columns with integer values and one ID.
df <- matrix(ncol=5,nrow = 10)
colnames(df) <- c("A","B","C","D","ID")
# df
for (i in 1:10){
df[i,1:4] <- sample(1:5,4, replace = TRUE)
}
df <- data.frame(df)
df$ID <- make.unique(rep(LETTERS,length.out=10),sep='')
df
A B C D ID
1 2 4 3 5 A
2 5 1 3 5 B
3 3 3 5 3 C
4 4 3 1 5 D
5 2 1 2 5 E
6 5 4 4 5 F
7 4 4 3 3 G
8 2 1 5 5 H
9 4 4 1 3 I
10 4 2 2 2 J
Second data frame has manual input, it's user input, I want to turn it into shiny app later on, that's why also I'm asking for optimization, because my code doesn't seem very neat to me.
df.man <- data.frame(matrix(ncol=5,nrow=1))
colnames(df.man) <- c("A","B","C","D","ID")
df.man$ID <- c("man")
df.man$A <- 4
df.man$B <- 4
df.man$C <- 3
df.man$D <- 4
df.man
A B C D ID
4 4 3 4 man
I want to filter rows from reference sequentially, following the rules:
If there is exact match in a whole row between reference table and manual than extract this(those) from reference and show me that row, if not then reduce number of matching columns from right to left until there is a match but not between less then two variables(columns A,B).
So with my limited knowledge I've wrote this:
# subtraction manual from reference
df <- df %>% dplyr::mutate(Adiff=A-df.man$A)%>%
dplyr::mutate(Bdiff=B-df.man$B)%>%
dplyr::mutate(Cdiff=C-df.man$C) %>%
dplyr::mutate(Ddiff=D-df.man$D)
# check manually how much in a row has zero difference and filter those
ifelse(nrow(df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0 & Ddiff==0)) != 0,
df0<-df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0 & Ddiff==0),
ifelse(nrow(df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0)) != 0,
df0<-df%>%filter(Adiff==0 & Bdiff==0 & Cdiff==0),
ifelse(nrow(df%>%filter(Adiff==0 & Bdiff==0)) != 0,
df0<-df%>%filter(Adiff==0 & Bdiff==0),
"less then two exact match")
))
tbl_df(df0[,1:5])
# A tibble: 1 x 5
A B C D ID
<int> <int> <int> <int> <chr>
1 4 4 3 3 G
It works and found ID G but looks ugly to me. So the first question is - What would be recommended way to improve this? Are there any functions, packages or smth I'm missing?
Second question - I want to complicate condition.
Imagine we have reference data set.
A B C D ID
2 4 3 5 A
5 1 3 5 B
3 3 5 3 C
4 3 1 5 D
2 1 2 5 E
5 4 4 5 F
4 4 3 3 G
2 1 5 5 H
4 4 1 3 I
4 2 2 2 J
Manual input is
A B C D ID
4 4 2 2 man
Filtering rules should be following:
If there is exact match in a whole row between reference table and manual than extract this(those) from reference and show me that row, if not then reduce number of matching columns from right to left until there is a match but not between less then two variables(columns A,B).
From those rows where I have only two variable matches filter those which has ± 1 difference in columns to the right. So I should have filtered case G and I from reference table from the example above.
keep going the way I did above, I would do the following:
ifelse(nrow(df0%>%filter(Cdiff %in% (-1:1) & Ddiff %in% (-1:1)))>0,
df01 <- df0%>%filter(Cdiff %in% (-1:1) & Ddiff %in% (-1:1)),
ifelse(nrow(df0%>%filter(Cdiff %in% (-1:1)))>0,
df01<- df0%>%filter(Cdiff %in% (-1:1)),
"NA"))
It will be about 11 columns at the end, but I assume it doesn't matter so much.
Keeping in mind this objective - how would you suggest to proceed?
Thanks!
This is a lot to sort through, but I have some ideas that might be helpful.
First, you could keep your df a matrix, and use row names for your letters. Something like:
set.seed(2)
df
A B C D
A 5 1 5 1
B 4 5 1 2
C 3 1 3 2
D 3 1 1 4
E 3 1 5 3
F 1 5 5 2
G 2 3 4 3
H 1 1 5 1
I 2 4 5 5
J 4 2 5 5
And for demonstration, you could use a vector for manual as this is input:
# Complete match example
vec.man <- c(3, 1, 5, 3)
To check for complete matches between the manual input and reference (all 4 columns), with all numbers, you can do:
df[apply(df, 1, function(x) all(x == vec.man)), ]
A B C D
3 1 5 3
If you don't have a complete match, would calculate differences between df and vec.man:
# Change example vec.man
vec.man <- c(3, 1, 5, 2)
df.diff <- sweep(df, 2, vec.man)
A B C D
A 2 0 0 -1
B 1 4 -4 0
C 0 0 -2 0
D 0 0 -4 2
E 0 0 0 1
F -2 4 0 0
G -1 2 -1 1
H -2 0 0 -1
I -1 3 0 3
J 1 1 0 3
The diffs that start with and continue with 0 will be your best matches (same as looking from right to left iteratively). Then, your best match is the column of the first non-zero element in each row:
df.best <- apply(df.diff, 1, function(x) which(x!=0)[1])
A B C D E F G H I J
1 1 3 3 4 1 1 1 1 1
You can see that the best match is E which was non-zero in the 4th column (last column did not match). You can extract rows that have 4 in df.best as your best matches:
df.match <- df[which(df.best == max(df.best, na.rm = T)), ]
A B C D
3 1 5 3
Finally, if you want all the rows with closest match +/- 1 if only 2 match, you could check for number of best matches (should be 3). Then, compare differences with vector c(0,0,1) which would imply 2 matches then 3rd column off by +/- 1:
# Example vec.man with only 2 matches
vec.man <- c(3, 1, 6, 9)
> df.match
A B C D
C 3 1 3 2
D 3 1 1 4
E 3 1 5 3
if (max(df.best, na.rm = T) == 3) {
vec.alt = c(0, 0, 1)
df[apply(df.diff[,1:3], 1, function(x) all(abs(x) == vec.alt)), ]
}
A B C D
3 1 5 3
This should be scalable for 11 columns and 4 matches.
To generalize for different numbers of columns, #IlyaT suggested:
n.cols <- max(df.best, na.rm=TRUE)
vec.alt <- c(rep(0, each=n.cols-1), 1)
Let's say I have this simple data frame:
df <- data.frame(x=c(1,3,3,1,3,1), y = c(2,2,2,2,2,2),z = c('a','b','c','d','e','f'))
> df
x y z
1 1 2 a
2 3 2 b
3 3 2 c
4 1 2 d
5 3 2 e
6 1 2 f
I would like to subset where x= 3, return only column x and y and include a calculated colum x+y.
I can get the first 2 things done, but I can't get the caclulated column to also appear.
df[df$x==3,c("x","y")]
How can I do that, but using base R only.
Staying in base, just do a rowSums before your subset.
df$xy <- rowSums(df[, c("x", "y")])
df[df$x == 3, c("x", "y", "xy")]
# x y xy
# 2 3 2 5
# 3 3 2 5
# 5 3 2 5
Personally, I do prefer the dplyr approach, which #akrun commented on your question.
You can also do like this
df <- data.frame(x=c(1,3,3,1,3,1), y = c(2,2,2,2,2,2),z = c('a','b','c','d','e','f'))
df$z <- ifelse(df$x == 3, (df$x + df$y), df$y)
df
x y z
1 1 2 2
2 3 2 5
3 3 2 5
4 1 2 2
5 3 2 5
6 1 2 2
I have a list of events and sequences. I would like to print the sequences in a separate table if event = x is included somewhere in the sequence. See table below:
Event Sequence
1 a 1
2 a 1
3 x 1
4 a 2
5 a 2
6 a 3
7 a 3
8 x 3
9 a 4
10 a 4
In this case I would like a new table that includes only the sequences where Event=x was included:
Event Sequence
1 a 1
2 a 1
3 x 1
4 a 3
5 a 3
6 x 3
Base R solution:
d[d$Sequence %in% d$Sequence[d$Event == "x"], ]
Event Sequence
1: a 1
2: a 1
3: x 1
4: a 3
5: a 3
6: x 3
data.table solution:
library(data.table)
setDT(d)[Sequence %in% Sequence[Event == "x"]]
As you can see syntax/logic is quite similar between these two solutions:
Find event's that are equal to x
Extract their Sequence
Subset table according to specified Sequence
We can use dplyr to group the data and filter the sequence with any "x" in it.
library(dplyr)
df2 <- df %>%
group_by(Sequence) %>%
filter(any(Event %in% "x")) %>%
ungroup()
df2
# A tibble: 6 x 2
Event Sequence
<chr> <int>
1 a 1
2 a 1
3 x 1
4 a 3
5 a 3
6 x 3
DATA
df <- read.table(text = " Event Sequence
1 a 1
2 a 1
3 x 1
4 a 2
5 a 2
6 a 3
7 a 3
8 x 3
9 a 4
10 a 4",
header = TRUE, stringsAsFactors = FALSE)
I am wanting to write a function so that a (potentially large) dataframe can be subsetted according to group membership, where a 'group' is a unique combination of a set of column values.
For example, I would like to subset the following data frame according to unique combination of the first two columns (Loc1 and Loc2).
Loc1 <- c("A","A","A","A","B","B","B")
Loc2 <- c("a","a","b","b","a","a","b")
Dat1 <- c(1,1,1,1,1,1,1)
Dat2 <- c(1,2,1,2,1,2,2)
Dat3 <- c(2,2,4,4,6,5,3)
DF=data.frame(Loc1,Loc2,Dat1,Dat2,Dat3)
Loc1 Loc2 Dat1 Dat2 Dat3
1 A a 1 1 2
2 A a 1 2 2
3 A b 1 1 4
4 A b 1 2 4
5 B a 1 1 6
6 B a 1 2 5
7 B b 1 2 3
I want to return (i) the number of groups (i.e. 4), (ii) the number in each group (i.e. c(2,2,2,1), and (iii) to relabel the rows so that I can further analyse the data frame according to group membership (e.g. for ANOVA and MANOVA) (i.e.
Group<-as.factor(c(1,1,2,2,3,3,4))
Data <- cbind(Group,DF[,-1:-2])
Group Dat1 Dat2 Dat3
1 1 1 1 2
2 1 1 2 2
3 2 1 1 4
4 2 1 2 4
5 3 1 1 6
6 3 1 2 5
7 4 1 2 3
).
So far all I have managed is to get the number of groups, and I'm suspicious that there's a better way to do even this:
nrow(unique(DF[,1:2]))
I was hoping to avoid for-loops as I am concerned about the function being slow.
I have tried converting to a data matrix so that I could concatenate the row values but I couldn't get that to work either.
Many thanks
You could try:
Create Group column by using unique level combination of Loc1 and Loc2.
indx <- paste(DF[,1], DF[,2])
DF$Group <- as.numeric(factor(indx, unique(indx))) #query No (iii)
DF1 <- DF[-(1:2)][,c(4,1:3)]
# Group Dat1 Dat2 Dat3
#1 1 1 1 2
#2 1 1 2 2
#3 2 1 1 4
#4 2 1 2 4
#5 3 1 1 6
#6 3 1 2 5
#7 4 1 2 3
table(DF$Group) #(No. ii)
#1 2 3 4
#2 2 2 1
length(unique(DF$Group)) #(i)
#[1] 4
Then, if you need to subset the datasets by group, you could split the dataset using the Group to create a list of 4 list elements
split(DF1, DF1$Group)
Update
If you have multiple columns, you could still try:
ColstoGroup <- 1:2
indx <- apply(DF[,ColstoGroup], 1, paste, collapse="")
as.numeric(factor(indx, unique(indx)))
#[1] 1 1 2 2 3 3 4
You could create a function;
fun1 <- function(dat, GroupCols){
FactGroup <- dat[, GroupCols]
if(length(GroupCols)==1){
dat$Group <- as.numeric(factor(FactGroup, levels=unique(FactGroup)))
}
else {
indx <- apply(FactGroup, 1, paste, collapse="")
dat$Group <- as.numeric(factor(indx, unique(indx)))
}
dat
}
fun1(DF, "Loc1")
fun1(DF, c("Loc1", "Loc2"))
This gets all three of your queries.
Begin with a table of the first two columns and then work with that data.
> (tab <- table(DF$Loc1, DF$Loc2))
#
# a b
# A 2 2
# B 2 1
#
> (ct <- c(tab)) ## (ii)
# [1] 2 2 2 1
> length(unlist(dimnames(tab))) ## (i)
# [1] 4
> cbind(Group = rep(seq_along(ct), ct), DF[-c(1,2)]) ## (iii)
# Group Dat1 Dat2 Dat3
# 1 1 1 1 2
# 2 1 1 2 2
# 3 2 1 1 4
# 4 2 1 2 4
# 5 3 1 1 6
# 6 3 1 2 5
# 7 4 1 2 3
Borrowing a bit from this answer and using some dplyr idioms:
library(dplyr)
Loc1 <- c("A","A","A","A","B","B","B")
Loc2 <- c("a","a","b","b","a","a","b")
Dat1 <- c(1,1,1,1,1,1,1)
Dat2 <- c(1,2,1,2,1,2,2)
Dat3 <- c(2,2,4,4,6,5,3)
DF <- data.frame(Loc1, Loc2, Dat1, Dat2, Dat3)
emitID <- local({
idCounter <- -1L
function(){
idCounter <<- idCounter + 1L
}
})
DF %>% group_by(Loc1, Loc2) %>% mutate(Group=emitID())
## Loc1 Loc2 Dat1 Dat2 Dat3 Group
## 1 A a 1 1 2 0
## 2 A a 1 2 2 0
## 3 A b 1 1 4 1
## 4 A b 1 2 4 1
## 5 B a 1 1 6 2
## 6 B a 1 2 5 2
## 7 B b 1 2 3 3